I need to prove:
lim (x^4)-(2(x^2)) = +inf
x->+inf
Any suggestions?
You want to show that for every $\displaystyle M > 0$, there exists $\displaystyle N > 0$ such that whenever $\displaystyle x > N$, it necessarily follows that $\displaystyle f(x) > M \Leftrightarrow x^4 - 2x^2 > M $
Consider the latter expression as an equality:
$\displaystyle x^4 - 2x^2 = M $
$\displaystyle \Leftrightarrow x^4 - 2x^2 - M = 0$
$\displaystyle \Leftrightarrow x^2 = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-M)}}{2}$
$\displaystyle \Leftrightarrow x^2 = \frac{2 \pm \sqrt{4 + 4M}}{2}$
$\displaystyle \Leftrightarrow x^2 = 1 \pm \sqrt{1 + M} $
$\displaystyle \Leftrightarrow x^2 = 1 + \sqrt{1 + M}$ (since $\displaystyle x^2 \geq 0$ and $\displaystyle M > 0$)
$\displaystyle \Leftrightarrow x = +\sqrt{1 + \sqrt{1 + M}}$ (we're considering positive x's)
So when $\displaystyle x \ {\color{red}=} \ \sqrt{1 + \sqrt{1 + M}}$ we have that $\displaystyle f(x)\ {\color{red} =}\ M$. We want to show that for some $\displaystyle x \ {\color{magenta}>}\ N$, we have that $\displaystyle f(x) \ {\color{magenta}>}\ M$
Now consider: $\displaystyle x \ {\color{magenta}> }\ \sqrt{1 + \sqrt{1 + M}} > 0$. Can you finish?